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% Created 2023-02-20 Mon 22:30
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\notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry}
\author{Lizzy Hunt}
\date{\today}
\title{Midterm One}
\hypersetup{
pdfauthor={Lizzy Hunt},
pdftitle={Midterm One},
pdfkeywords={},
pdfsubject={},
pdfcreator={Emacs 28.2 (Org mode 9.6.1)},
pdflang={English}}
\begin{document}
\maketitle
\setlength\parindent{0pt}
\section{Question One}
\label{sec:org6315799}
\begin{verbatim}
In [1]: sols = lambda n: set(filter(lambda x: (x**2 + x) % n == 0, range(n)))
In [2]: sols(5)
Out[2]: {0, 4}
In [3]: sols(6)
Out[3]: {0, 2, 3, 5}
\end{verbatim}
\section{Question Two - TODO}
\label{sec:org6c24b4c}
\section{Question Three}
\label{sec:org76bcd7b}
From the multiplication table we can see that \(e\) is the identity element \(1_R\), the zero
element \(0_R\) is \(z\). From the addition table we find that every element is its own
additive inverse.
\subsection{a}
\label{sec:org983fdae}
The units in this ring are \({e, g}\), and the zero divisors are \({a, b, c, f}\).
\subsection{b}
\label{sec:org9072029}
\(3bd = 3(bd) = 3z = b\)
\(2ac = 2(ac) = 2b = b + b = z\)
\(g^{-1}f^3 = (g)(fff) (gf)(ff) = (d)(f) = d\)
Thus, \(3bd - 2ac + g^{-1}f^3 = b - z + d = b + d = f\).
\section{Question Four}
\label{sec:orge571a7e}
To get a hunch,
\begin{verbatim}
In [1]: import numpy as np
In [2]: def find_counter(matrix_generator, in_ring, search_space=range(-100, 100)):
...: for a1 in search_space:
...: n = matrix_generator(a1)
...: for a2 in search_space:
...: m = matrix_generator(a2)
...: dot = np.dot(n, m)
...: add = np.add(n, m)
...:
...: if not in_ring(dot):
...: return (n, m, dot, '*')
...: if not in_ring(add):
...: return (n, m, add, '+')
...:
...: return None
...:
In [3]: find_counter(lambda a: [[0, a], [0, -a]], lambda m: m[0][1] == -m[1][1] and m[0][0] == 0 and m[1][0] == 0)
In [4]: In [31]: find_counter(lambda a: [[0, a], [a, 0]], lambda m: m[0][1] == m[1][0] and m[1][1] == 0 and m[0][0] == 0)
Out[4]:
([[0, -100], [-100, 0]],
[[0, -100], [-100, 0]],
array([[10000, 0],
[ 0, 10000]]),
'*')
\end{verbatim}
\subsection{a}
\label{sec:orge0ecef4}
Using Theorem 3.2, this is a ring:
\begin{enumerate}
\item S\textsubscript{1} is closed under addition: \(x, y \in S_1 \Rightarrow x + y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x + y = \begin{smallmatrix} 0 & a + b \\ 0 & (-a) + (-b) \end{smallmatrix} = \begin{smallmatrix} 0 & a + b \\ 0 & -(a + b) \end{smallmatrix}\) which satisfies the rule
\item S\textsubscript{1} is closed under multiplication: \(x, y \in S_1 \Rightarrow x \cdot y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x \cdot y =\) \(\begin{smallmatrix} 0 + (a)(0) & 0(b) + a(-b) \\ 0 + (-a)(0) & 0b + (-a)(-b) \end{smallmatrix}\) \(=\) \(\begin{smallmatrix} 0 & -ab \\ 0 & ab \end{smallmatrix}\) which satisfies the rule
\item \(0_R_{} = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\)
\item \(a \in S_1 \Rightarrow a + x = 0_R \wedge x \in S_1\) since \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(x = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) \(\Rightarrow x =\) \(\begin{smallmatrix} 0 & -a \\ 0 & a \end{smallmatrix}\) \(\in S_1\)
\end{enumerate}
\subsection{b}
\label{sec:org1a8af9b}
\(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} 10000 & 0 \\ 0 & 10000 \end{smallmatrix}\) is not in the ring.
\section{Question Five}
\label{sec:org85339f8}
No, it is not a homomorphism from the definition in 3.3.
Consider \(n =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix}\) and \(m =\) \(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\).
Then \(nm =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) and thus \(Trace(n) \cdot Trace(m) = 1 \wedge Trace(nm) = 0 \Rightarrow f(n)f(m) \neq f(nm)\).
\section{Question Six}
\label{sec:org405929b}
No, I don't believe \(f\) to be an isomorphism.
\subsection{Lemma}
\label{sec:org323f970}
We need to show that \(x \in Q \wedge x \sqrt{2} = y \in Q \Rightarrow x = 0\).
If \(x \neq 0\) then \(y\) can be represented as \(\frac{a}{b}\) with \(a, b \neq 0 \wedge a,b \in \mathds{Z}\) and \((a, b) = 1\).
So, \(x \sqrt{2} = \frac{a}{b} \Rightarrow x \sqrt{2} b = a \Rightarrow x^2(2b^2) = a^2\) and thus \(2 | aa\), and by applying Corollary 1.6, \(2 | a\).
By substituting \(2c = a\) (a is divisible by 2), \(x^2(2b^2) = 4c^2 \Rightarrow x^2 b^2 = 2c^2\) which implies \(2 | (x^2 b^2) \Rightarrow 2 | x\) or \(2 | b\) (again by Corollary 1.6).
But 2 cannot divide \(b\) as well as \(a\), since \((a, b) = 1\) and \(b \neq 0\).
Thus \(2 | x\). So \(x = 2d \Rightarrow 4d^2(2b^2) = a^2 \Rightarrow 4 | a^2\), so \(4e = a\), and by similar logic for \(b\) above, \(4 | x\).
Then, subsituting \(x = 4f \Rightarrow 16f^2(2b^2) = a^2 \Rightarrow 16 | a^2\). Again, \(16 | x\).
In fact, all range elements \(R\) of the function on the natural numbers \$f \(\ni\) \$ \(f(1) = 2, f(n) = f(n-1)^2 { n > 1 }\) must divide \(x\).
Thus, x must be sup(\(R\)) or 0. Obviously \(R\) is unbounded on the right, so it follows \(x = 0\).
\subsection{Proof}
\label{sec:orge723534}
\(a, b \in Q(\sqrt{2}) \Rightarrow f(a) = n + m \sqrt{2} \wedge b = x + y \sqrt{2} \Rightarrow f(a) = n - m \sqrt{2} \vee f(b) = x - y \sqrt{2}\).
\(f\) is not injective; if we assume \(f(a) = f(b) \Rightarrow a = b\) then \(f(a) - f(b) = 0 \Rightarrow n - m \sqrt{2} - x - y \sqrt{2} = 0 \Rightarrow 0 = (n-x) - (m + y)\sqrt{2} \Rightarrow n-x = (m+y)\sqrt{2}\) and, as \(n-x \in Q\) since
Q is a ring and by definition, \((m + y)\sqrt{2} \in Q\) implies \((m + y) = 0\) by the lemma, thus \(m = -y\). But for \(a = b\), \(m = y\), a contradiction.
\end{document}
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